- anonymous

OMG!!!!!!!!!!! HELP PLZZZZZZZZ

- chestercat

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- anonymous

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- anonymous

- anonymous

- anonymous

- anonymous

plzzzzz help

- anonymous

hello??????????

- Jack1

for this... u srsly need to flip a coin 50 times and fill in the table from ur results @bubbleslove1234 ... sorry, till u do we cant help u hey

- anonymous

i have already done this

- LunyMoony

yeah, I don't understand how they want the statistics either for the 2, or 2 each, of cats and dogs on the chart

- Jack1

so what were ur answers for:
HH -
HT -
TH -
and
TT? -

- anonymous

hold on

- anonymous

hh=11
ht=17
th=8
tt=14

- Jack1

ok... so from ur screenshot: which question do u want to go through first?

- anonymous

the first one and work from there

- Jack1

k_ so cats are Heads, and Dogs are Tails
according to the start of ur Q: having cat or dog is 50:50 "theoretical" probability
so: Q1 - theoretical Probability of 2 cats OR 2 dogs
= P(HH) OR P(TT)
= P(HH) + P(TT)
Theoretical probability of HH =0.25
Theoretical probability of TH =0.25
Theoretical probability of HT =0.25
Theoretical probability of TT =0.25
so solve
= P(HH) + P(TT)
@bubbleslove1234 ?

- anonymous

do you know what i should get cause i am soooooooooooo confused

- anonymous

oww my brain hurts

- Jack1

Theoretical probability is:
P(HH) =0.25 (or 25%)
P(TH) =0.25 (or 25%)
P(HT) =0.25 (or 25%)
P(TT) =0.25 (or 25%)
so Q1. solve
= P(HH) + P(TT)

- anonymous

okay so 25% times 50

- Jack1

no, sorry u can either use percentages or use decimals, but u can't use both at the same time, sorry
P(HH) = 0.25
P(TT) = 0.25
so = P(HH) + P(TT) =?

- anonymous

okay

- Jack1

P(HH) = 0.25
P(TT) = 0.25
so
P(HH) + P(TT) = ?????
@bubbleslove1234 ...?

- anonymous

p(11)=0.25
p(14)=0.25

- arindameducationusc

did you get the Question 1 answer? dog and cat?

- anonymous

nope not yet but i need fast help cause it is due today

- Jack1

no, for question 1, we only care about Theoretical results... not actual one we get from flipping a coin
so Theoretical probability of 2 heads is 25% of the time... so P(HH) = 0.25
and the Theoretical probability of 2 tails is 25% of the time... so P(TT) = 0.25
so whats .25 + .25 ?

- anonymous

50

- Jack1

perfect. PHH + PTT = 0.50 .... or 50 %
so Q1
= 50% chance

- anonymous

oh so is that the answer

- Jack1

Q1. chance family has 2 cats or 2 dogs?
= P(CC) or P(DD)
= 0.5 .... or 50%
yep, answer

- anonymous

for the first one

- Jack1

Q2.
how does 2 coins simulate which 2 pets a family has?
answer: 50% chance of choosing a cat or a dog, and they chose twice
(50% either cat or dog, then 50% chance either cat or dog for the 2nd pet also)
so exactly as flipping a coin... twice
(50% either heads or tail, then 50% chance either head or tail on the 2nd toss also)

- Jack1

Q3.check if this is correct:
ur results were:
|dw:1439454960600:dw|

- Jack1

|dw:1439455108667:dw|

- Jack1

Q4.
Experimental Probability
In ur test, how many times did u flip the coin? @bubbleslove1234 ?

- anonymous

50

- Jack1

perfect
now, calculate your results for P(HH) and P(TT)
P(HH) = number of results that were HH / total number of flips
=11/50
= what? (in decimals)
P(TT) = number of results that were *TT / total number of flips
=14/50
= what? (in decimals) @bubbleslove1234 ?

- Jack1

sorry, typo

- anonymous

P(HH) = 0.22
P(TT) =0.28

- Jack1

absolutely right
so: Experimental Probability of 2 cats or 2 dogs = same probability as flipping a coin 50 times and getting 2 heads or 2 tails
so = P(HH) + P(TT)
= 0.22 + 0.28
= ???

- anonymous

0.5

- anonymous

- arindameducationusc

@bubbleslove1234
should I try?

- anonymous

sure

- arindameducationusc

okay for question 1,
Dog/Cat
1 1
2 0
0 2
now,
in the question its given 2 cats or 2 dogs, so
trials are 3, so, its 2/3

- Jack1

sorry my page froze... and is being glitchy
can u finish @arindameducationusc please?

- arindameducationusc

okay @Jack1
let me try.........

- arindameducationusc

For question 2, just assign head as dog and tail as cat

- anonymous

kk

- arindameducationusc

now,
hh=11
ht=17
th=8
tt=14
if this your data.....
then count head head+tail+tail (according to question4 2 dogs and 2 cats)
so, 11+14=25
now for probability divide by the frequency,
so answer is 25/50

- anonymous

okay

- anonymous

got it

- anonymous

can we move to Q5

- arindameducationusc

yes sure,

- anonymous

thanks

- arindameducationusc

for question 5,
C=Cat, D=Dog
CCD
CDD
DDD
DCC
DDC
CCC
CDC
DCD
now, count where CCC and DDD is there
which is 2.
2/8 , where 8 is the frequesncy.
2/8 is the probability

- Jack1

hey, back and working, but keep going, happy to watch
Q5: perfect, but 1 question:
how did u get 2/8 for Q5... but only 2/3 for Q1?
should it not be 2/4...?

- arindameducationusc

Q5 and Q1 has no relation (very important)

- arindameducationusc

one sec....

- Jack1

no.
categorically: no. Q1 and Q5 are totally related ;)

- arindameducationusc

Q5 is correct.....

- arindameducationusc

but Q1 I missed something.......

- anonymous

what

- arindameducationusc

if D=Dog
C=Cat
CC
DD
CD
DC
so yes it is 2/4.
Thank you @Jack1

- Jack1

np dude ;)

- arindameducationusc

Now question 6

- anonymous

kk

- arindameducationusc

H=Head, T=Tail
In the table given in Question, replace that with
TTH
THH
HHH
HTT
HHT
TTT
THT
HTH

- anonymous

???????

- anonymous

what

- Jack1

he means : flip a coin 3 times, and record the results
current simulation is; flip a coin 2 times

- anonymous

so what is the answer

- Jack1

##### 1 Attachment

- arindameducationusc

@Jack1
Exactly !
Flipping coins 3 times.... and record observations

- anonymous

only 3 times

- Jack1

flip a coin... 3 times.... record the results... repeat 50 times
or flip 3 coins at the same time... 50 times

- anonymous

okay

- anonymous

thanks

- arindameducationusc

Good job @bubbleslove1234 ...
Was my attempt useful @bubbleslove1234 ?

- anonymous

yep

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